MHB Solve Quadratic Equation: $\mathcal{E}$ Experiment

  • Thread starter Thread starter Julio1
  • Start date Start date
AI Thread Summary
The discussion centers on the conditions under which the quadratic equation \( ax^2 + bx + c = 0 \) has two distinct real roots, specifically focusing on the coefficients \( a, b, \) and \( c \). For the equation to have real roots, the coefficients must be real numbers, and the discriminant \( b^2 - 4ac \) must be greater than zero. The choice of coefficients significantly impacts the probability of obtaining distinct real roots, with different distributions yielding varying results. For example, if coefficients are integers from a dice throw, the probability can be calculated, while complex coefficients drawn from a uniform distribution lead to a zero-probability event for real roots. The conversation emphasizes the importance of understanding the probability distribution of the coefficients in determining the nature of the roots.
Julio1
Messages
66
Reaction score
0
The $\mathcal{E}$ experiment consists in choosing the numbers $a, b$ and $c$. Describe the event $A \,= \,$ the equation $ax^2+bx+c=0$ has two distinct real roots.
Hi !, I have this problem, I understand that the roots of the equation are $x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a},$ i.e., $x_1=\dfrac{-b+\sqrt{b^2-4ac}}{2a}$ y $x_2=\dfrac{-b-\sqrt{b^2-4ac}}{2a},$ but I don't understand that more there that do. :confused:
 
Mathematics news on Phys.org
How are the coefficients chosen? The answer is going to depend on that (for instance, if they are integers chosen from a dice throw the probability is going to be computable from basic probability, but if they are complex numbers drawn from a uniform distribution then $A$ is a zero-probability event).
 
Bacterius said:
How are the coefficients chosen? The answer is going to depend on that (for instance, if they are integers chosen from a dice throw the probability is going to be computable from basic probability, but if they are complex numbers drawn from a uniform distribution then $A$ is a zero-probability event).

Hello Bacterius, Thanks !

For that the equation $ax^2+bx+c=0$ has real roots, the coefficients $a,b$ and $c$ must be reals. In other hand, if $a,b,c\in \mathbb{C}$ the equation has complex roots, but how solve now the problem?
 
That wasn't what Bacterius meant when he asked how the coefficients are chosen. What probability distribution is used? It can't be "uniform" because there is no uniform distribution over the entire real numbers.

In any case, the equation ax^2+ bx+ c= 0 has two distinct real roots if and only if the discriminant, b^2- 4ac, is greater than 0. The probability of that depends upon the probability distribution of a, b, and c.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

Similar threads

Replies
6
Views
2K
Replies
1
Views
1K
Replies
16
Views
4K
Replies
19
Views
3K
Replies
1
Views
1K
Replies
4
Views
1K
Replies
4
Views
1K
Back
Top